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Year 13 · AQA & OCR A
Practise the calculations that usually cost marks: products minus reactants for ΔS°, matching J with kJ in ΔG = ΔH − TΔS, and deciding what the sign of ΔG actually tells you.
Multiply each standard molar entropy by its coefficient in the balanced equation. Add the product terms, add the reactant terms, then calculate products minus reactants.
ΔH is usually in kJ mol−1. ΔS is usually in J K−1 mol−1. Divide ΔS by 1000 before using it with ΔH in kJ mol−1.
A negative ΔG predicts that the forward change is thermodynamically feasible at the stated temperature. ΔG = 0 is the boundary. A positive ΔG gives an unfavourable thermodynamic result for the forward change under those conditions.
OCR A: reaction rate depends on the activation-energy barrier and available pathway. A reaction with a negative ΔG can still be extremely slow when the barrier is large.
